Answer
The solutions are $y=-3\pm\dfrac{\sqrt{30}}{2}$
Work Step by Step
$2y^{2}+12y+3=0$
Take $3$ to the right side:
$2y^{2}+12y=-3$
Take out common factor $2$ from the left side:
$2(y^{2}+6y)=-3$
Take $2$ to divide the right side:
$y^{2}+6y=-\dfrac{3}{2}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both side of the equation. In this particular case, $b=6$:
$y^{2}+6y+\Big(\dfrac{6}{2}\Big)^{2}=-\dfrac{3}{2}+\Big(\dfrac{6}{2}\Big)^{2}$
$y^{2}+6y+9=-\dfrac{3}{2}+9$
$y^{2}+6y+9=\dfrac{15}{2}$
Factor the left side of the equation, which is a perfect square trinomial:
$(y+3)^{2}=\dfrac{15}{2}$
Take the square root of both sides of the equation:
$\sqrt{(y+3)^{2}}=\pm\sqrt{\dfrac{15}{2}}$
$y+3=\pm\sqrt{\dfrac{15}{2}}$
Solve for $y$:
$y=-3\pm\sqrt{\dfrac{15}{2}}=-3\pm\dfrac{\sqrt{30}}{2}$
